A sample of normally consolidated clay was subjected to a CU triaxial compression test that was carried out until the specimen failed at a deviator stress of 50 kN/m2. The pore water pressure at failure was recorded to be 18 kN/m2, and a confining pressure of 48 kN/m2 was used in the test. Using any software available at UGA, draw both the effective stress strength envelope and total stress strength envelope obtained form this CU test. Based on unconfined compression test using same clay, 20 kN/m2 of unconfined compressive stress was measured at the failure.

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temmydbrain temmydbrain · Answer 1 · 2020-05-02T03:57:10+02:00

Answer:

Check the explanation

Explanation:

Given

1) CU traixial compression test

2) Devatoric stress at failure = бd = 50 kN/[tex]m^2[/tex]

3) Confining pressure at failure = бd = 48 kN/[tex]m^2[/tex]

4) Pore pressure at failure = u = 18 kN/[tex]m^2[/tex]

5) Unconfined compression stress = q = 20 kN/[tex]m^2[/tex]

6) Undrained cohesion = q/2 = 20 kN/[tex]m^2[/tex]

To find:

1) Effective and total stress strength failure envelope

Kindly check the attached image below .

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